Class 10 Mathematics Notes Chapter 6 (Triangles) – Mathematics Book

Here are the detailed notes on Chapter 6, 'Triangles', from your NCERT Class 10 Mathematics textbook. This is a very important chapter, not just for your board exams but also forms the basis for many concepts tested in various government examinations. We'll cover the core ideas, theorems, and problem-solving techniques. Pay close attention!

Chapter 6: Triangles - Detailed Notes for Government Exam Preparation

1. Introduction: Congruence vs. Similarity

• Congruent Figures: Figures having the same shape and the same size. Think of identical twins or two coins of the same denomination. (Covered in Class 9). Symbol: ≅
• Similar Figures: Figures having the same shape but not necessarily the same size. Think of a photograph and its enlargement, or different sized circles. Symbol: ~
  • All congruent figures are similar, but the reverse is not necessarily true.
  • For polygons to be similar:
    • Their corresponding angles must be equal.
    • Their corresponding sides must be in the same ratio (or proportion).

2. Similarity of Triangles

This is the core focus of the chapter. Two triangles are similar if:

• (i) Their corresponding angles are equal.
  If ΔABC ~ ΔPQR, then ∠A = ∠P, ∠B = ∠Q, ∠C = ∠R.
• (ii) Their corresponding sides are in the same ratio (or proportion).
  If ΔABC ~ ΔPQR, then AB/PQ = BC/QR = AC/PR. This common ratio is called the scale factor.

Important Note: If either of the above conditions holds true for two triangles, the other condition will automatically hold true. This leads to the criteria for similarity.

3. Criteria for Similarity of Triangles

We don't need to check all angles and all sides every time. We have specific criteria:

• (i) AAA (Angle-Angle-Angle) Similarity Criterion: If in two triangles, corresponding angles are equal, then their corresponding sides are in the same ratio (hence, the triangles are similar).

  • Corollary AA (Angle-Angle) Similarity: If two angles of one triangle are respectively equal to two angles of another triangle, then the two triangles are similar (because the third angle will automatically be equal due to the angle sum property). This is the most frequently used criterion.

• (ii) SSS (Side-Side-Side) Similarity Criterion: If in two triangles, the sides of one triangle are proportional to (i.e., in the same ratio of) the sides of the other triangle, then their corresponding angles are equal (and hence the two triangles are similar).

  • If AB/PQ = BC/QR = AC/PR, then ΔABC ~ ΔPQR.

• (iii) SAS (Side-Angle-Side) Similarity Criterion: If one angle of a triangle is equal to one angle of the other triangle and the sides including these angles are proportional, then the two triangles are similar.

  • If ∠A = ∠P and AB/PQ = AC/PR, then ΔABC ~ ΔPQR. (Note: The equal angle must be between the proportional sides).

4. Important Theorems

• Theorem 6.1: Basic Proportionality Theorem (BPT) or Thales' Theorem:

  • Statement: If a line is drawn parallel to one side of a triangle intersecting the other two sides in distinct points, then the other two sides are divided in the same ratio.
  • In ΔABC, if DE || BC (where D is on AB and E is on AC), then AD/DB = AE/EC.
  • Corollaries (also very useful):
    • AD/AB = AE/AC
    • DB/AB = EC/AC

• Theorem 6.2: Converse of Basic Proportionality Theorem:

  • Statement: If a line divides any two sides of a triangle in the same ratio, then the line is parallel to the third side.
  • In ΔABC, if AD/DB = AE/EC, then DE || BC.

• Theorem 6.6: Ratio of Areas of Similar Triangles:

  • Statement: The ratio of the areas of two similar triangles is equal to the square of the ratio of their corresponding sides.
  • If ΔABC ~ ΔPQR, then Area(ΔABC) / Area(ΔPQR) = (AB/PQ)² = (BC/QR)² = (AC/PR)²
  • Important Extension: This ratio is also equal to the square of the ratio of their corresponding:
    • Altitudes
    • Medians
    • Angle Bisectors
    • Perimeters (Ratio of perimeters is equal to the ratio of sides, not squared). i.e., Perimeter(ΔABC) / Perimeter(ΔPQR) = AB/PQ.

• Theorem 6.8: Pythagoras Theorem:

  • Statement: In a right-angled triangle, the square of the hypotenuse is equal to the sum of the squares of the other two sides.
  • In ΔABC, if ∠B = 90°, then AC² = AB² + BC².
  • (Proof using similar triangles is part of this chapter's syllabus): Draw BD ⊥ AC. Prove ΔADB ~ ΔABC and ΔBDC ~ ΔABC. Add results.

• Theorem 6.9: Converse of Pythagoras Theorem:

  • Statement: In a triangle, if the square of one side is equal to the sum of the squares of the other two sides, then the angle opposite the first side is a right angle.
  • In ΔABC, if AC² = AB² + BC², then ∠B = 90°.

5. Problem-Solving Tips for Exams:

• Identify Similarity: Look for parallel lines (suggests BPT or AA similarity using corresponding/alternate angles), right angles (suggests Pythagoras or similarity in right triangles), or given ratios of sides.
• State the Criterion: When proving similarity, always explicitly state the criterion used (AA, SSS, SAS).
• Match Corresponding Parts: Be very careful while writing the similarity relation (e.g., ΔABC ~ ΔPQR). The order of vertices matters! A corresponds to P, B to Q, C to R. This helps in setting up the correct ratios of sides.
• Use Theorems Appropriately: Know when to apply BPT (parallel line dividing sides), Area Theorem (ratio of areas needed), or Pythagoras (right-angled triangle).
• Diagrams: Draw clear diagrams, even if not provided. Mark known angles and side lengths.

This chapter requires practice. Work through the NCERT examples and exercises thoroughly. Focus on understanding the proofs of the major theorems, especially BPT and Pythagoras using similarity, as conceptual understanding is often tested.

Multiple Choice Questions (MCQs)

Here are 10 MCQs based on Chapter 6 concepts, suitable for government exam practice:

1. In ΔABC, DE || BC such that D is on AB and E is on AC. If AD = 2.4 cm, AE = 3.2 cm, and EC = 4.8 cm, then AB is equal to:
   (a) 3.6 cm
   (b) 6 cm
   (c) 6.4 cm
   (d) 7.2 cm

2. If ΔABC ~ ΔPQR, Area(ΔABC) = 81 cm² and Area(ΔPQR) = 144 cm². If BC = 6 cm, then QR is equal to:
   (a) 8 cm
   (b) 9 cm
   (c) 12 cm
   (d) 10 cm

3. Which of the following is NOT a criterion for similarity of triangles?
   (a) AAA
   (b) SAS
   (c) ASA
   (d) SSS

4. In ΔLMN, ∠L = 50°, ∠M = 60°. If ΔLMN ~ ΔXYZ, then ∠Z is equal to:
   (a) 50°
   (b) 60°
   (c) 70°
   (d) 110°

5. Sides of two similar triangles are in the ratio 4 : 9. The ratio of the areas of these triangles is:
   (a) 2 : 3
   (b) 4 : 9
   (c) 81 : 16
   (d) 16 : 81

6. In ΔDEF, a line intersects DE at P and DF at Q such that DP/PE = DQ/QF. Which of the following must be true?
   (a) PQ ⊥ DF
   (b) PQ || EF
   (c) PQ = 1/2 EF
   (d) ∠DPQ = ∠DFE

7. A vertical pole of length 6 m casts a shadow 4 m long on the ground. At the same time, a tower casts a shadow 28 m long. The height of the tower is:
   (a) 40 m
   (b) 42 m
   (c) 36 m
   (d) 45 m

8. In an isosceles triangle ABC, if AC = BC and AB² = 2AC², then ∠C is:
   (a) 45°
   (b) 60°
   (c) 90°
   (d) 30°

9. If ΔABC ~ ΔQRP, Area(ΔABC)/Area(ΔPQR) = 9/4, AB = 18 cm and BC = 15 cm, then PR is equal to:
   (a) 10 cm
   (b) 12 cm
   (c) 20/3 cm
   (d) 8 cm

10. In a right-angled triangle PQR, right-angled at Q, QS is the altitude to the hypotenuse PR. Which of the following is TRUE?
    (a) ΔPSQ ~ ΔRSQ
    (b) ΔPQS ~ ΔPRQ
    (c) ΔRQS ~ ΔRPQ
    (d) Both (b) and (c)

Answer Key for MCQs:

1. (b) 6 cm (Use BPT: AD/DB = AE/EC => 2.4/DB = 3.2/4.8 => DB = 3.6 cm. AB = AD + DB = 2.4 + 3.6 = 6 cm)
2. (a) 8 cm (Use Area Theorem: Area(ABC)/Area(PQR) = (BC/QR)² => 81/144 = (6/QR)² => 9/12 = 6/QR => QR = (6 * 12) / 9 = 8 cm)
3. (c) ASA (ASA is a criterion for congruence, not similarity. AA is sufficient for similarity).
4. (c) 70° (In ΔLMN, ∠N = 180° - 50° - 60° = 70°. Since ΔLMN ~ ΔXYZ, corresponding angles are equal. ∠Z corresponds to ∠N, so ∠Z = 70°).
5. (d) 16 : 81 (Ratio of areas = Square of ratio of corresponding sides = (4/9)² = 16/81).
6. (b) PQ || EF (This is the Converse of Basic Proportionality Theorem).
7. (b) 42 m (The sun's rays create similar triangles. Let height of tower be h. Height/Shadow ratio is constant. 6/4 = h/28 => h = (6 * 28) / 4 = 42 m).
8. (c) 90° (Given AB² = 2AC². Since AC = BC, we can write AB² = AC² + AC² = AC² + BC². This matches the Pythagoras theorem format a² + b² = c², with AB as the hypotenuse. The angle opposite the hypotenuse AB is ∠C, so ∠C = 90°).
9. (a) 10 cm (Careful with correspondence: ΔABC ~ ΔQRP. Area(ABC)/Area(QRP) = (BC/RP)² => 9/4 = (15/PR)² => 3/2 = 15/PR => PR = (15 * 2) / 3 = 10 cm).
10. (d) Both (b) and (c) (When an altitude is drawn from the right angle vertex to the hypotenuse, it divides the triangle into two smaller triangles that are similar to the original triangle and to each other. So, ΔPQS ~ ΔPRQ and ΔRQS ~ ΔRPQ (using AA similarity)).

Make sure you understand the reasoning behind each answer. Good luck with your preparation!